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3 years ago

What is the solution of the following system

Mathematics

1 answer:

AleksAgata [21]3 years ago

4 0

Step-by-step explanation:

8x + 14y = 4 ...1

-6x - 7y = -10 ...2

...2 × 2 in 2 sides of the equation

-12x -14y = -20 ...3

...1 + ...3

8x + 14y +(-12x -14y) = 4 +(-20)

8x + 14y - 12x -14y = 4 - 20

-4x = -16

x = (-16)/(-4)

= 4

replace x in ...1 or ...2 you will reserve y value

and answer in the term of (x,y)

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=2x2+17x+35

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Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they

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The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

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$n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276$

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$n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220$

The probability that all are male of choosing '3' students

$P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }$

$P(E) = \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}$

P(E) = 0.067 = 6.71%

<u><em>Final answer</em></u>:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

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